If $Ax=b$ and $Ay=b$, how do I show $A((1-t)x+ty)=b$ for all $t\in \mathbb R$?

Question

If $Ax=b$ and $Ay=b$, how do I show $A((1-t)x+ty)=b$, $t\in \mathbb R$ ? How do I show that if a system of linear equations has more than one solution then it has infinitely many solutions?

Answer 1

$Ax = b = Ay, \tag 1$

so

$A((1 - t)x + ty) = (1 - t)Ax + tAy = (1 - t)b + tb = b - tb + tb = b. \tag 2$

This shows that if the system(s) (1) has more than one solution, i.e., $x \ne y$, then it has a continuum of solutions.

Note Added in Edit, Thursday 22 March 2018 9:18 AM PST: This note is in response to our OP Daniel Troncoso's question posted in a comment to this answer, asking why this answer shows the system has infinitely many solutions. Given the two solutions $x \ne y$ of (1), we first look at the case $x$ and$y$ linearly independent; if they are, the the map

$t \to (1 - t)x + ty, \; t \in \Bbb R, \tag 3$

is injective, since if for $t_1, t_2 \in \Bbb R$ we have

$(1 - t_1)x + t_1 y = (1 - t_2)x + t_2 y; \tag 4$

then

$(t_2 - t_1)x + (t_1 - t_2)y = 0; \tag 5$

the linear independence of $x$ and $y$ now forces

$t_2 - t_1 = 0 \Longrightarrow t_1 = t_2, \tag 6$

establishing the injectivity of (3); this means each $t \in \Bbb R$ maps to a unique $(1 - t)x + ty$, so the set

$\{(1 - t)x + ty \mid t \in \Bbb R \} \tag 7$

has the same cardinality as $\Bbb R$, and there are an uncountable infinity of solutions in this case. Now suppose $x$ and $y$ are linearly dependent; we break this case down into subcases according to whether one of $x$, $y$ is zero or not. If, say $x = 0$, then $y \ne 0$ since by assumption $x \ne y$; then the map

$t \to (1 - t) x + ty = ty, \tag 8$

and it is easy to see this is injective. Also, from $x = 0$ we have

$b = Ax = A(0) = 0, \tag 9$

so

$Ay = b = 0, \tag{10}$

whence

$A(ty) = tAy = 0, \tag{11}$

which gives us another way of seeing that the set $\{ty \mid t \in \Bbb R \}$ consists of solutions to (1), and that there are thus an uncountable infinity of same. Of course, the situation $y = 0 \ne x$ can be handled in the same fashion, with the same conclusion, so we are left with the case $x \ne 0 \ne y$; since $x$ and $y$ are assumed linearly dependent, there are $\alpha, \beta \in \Bbb R$, not both zero, with

$\alpha x + \beta y = 0; \tag{12}$

in fact, we must have $\alpha \ne 0 \ne \beta$, since if, say $\beta = 0$ then

$\alpha x = 0, \tag{13}$

which implies, since $\alpha \ne 0$,

$x = 0, \tag{14}$

a contradiction. If instead we had taken $\alpha = 0 \ne \beta$, we would obtain the also prohibited result $y = 0$; therefore we must have $\alpha \ne 0 \ne \beta$ and (12) may be written

$y = -\dfrac{\alpha}{\beta} x = c x, \; c \ne 0, \tag{15}$

that is, $x$ and $y$ are co-linear. The map (3) may now be written

$t \to (1 - t)x + ty = (1 - t)x + tcx = (1 - t + tc)x = (1 + (c - 1)t)x; \tag{16}$

then

$(1 + (c -1)t_1)x = (1 + (c - 1)t_2)x \Longrightarrow x + (c - 1)t_1 x = x + (c - 1)t_2 x$ $\Longrightarrow (c - 1) t_1 x = (c -1) t_2 x \Longrightarrow t_1x = t_2 x $ $\Longrightarrow (t_1 - t_2) x = 0 \Longrightarrow t_1 - t_2 = 0 \Longrightarrow t_1 = t_2, \tag{17}$

which shows that the map (3) is injective when $x$ and $y$ are linearly dependent as well; thus the solution set

$\{(1 + (c - 1)t)x \} \tag{18}$

has the cardinality of the continuum. We have thus covered all cases, and in each one there is a one-one mapping 'twixt $\Bbb R$ and the space of solutions, which are then infinite in number. End of Note.

Answer 2

If you have $Ax=b$ and $Ay=b$ you obviously have $Ax -A y = b-b=0$. This reduces to $A(x-y)=0$. Let's say $q:=(x-y)$.

Since this is a linear system, you can scale $q$ with any real number you like, to get an equally scaled solution on the right hand side. Therefor it holds $A\cdot(t q) =0$. Therefore you have $Ax + A(tq) =b+0 \Leftrightarrow A(x+tq)=b$. This holds for all possible values of $t$, therefore infitly many.